I What is the outer product of a tensor product of vectors?

If one has two single-particle Hilbert spaces ##\mathcal{H}_{1}## and ##\mathcal{H}_{2}##, such that their tensor product ##\mathcal{H}_{1}\otimes\mathcal{H}_{2}## yields a two-particle Hilbert space in which the state vectors are defined as $$\lvert\psi ,\phi\rangle =\lvert\psi\rangle\otimes\lvert\phi\rangle\in\mathcal{H}_{1}\otimes\mathcal{H}_{2}$$ where ##\lvert\psi\rangle\in\mathcal{H}_{1}## and ##\lvert\phi\rangle\in\mathcal{H}_{2}##.

Now, the inner product for ##\mathcal{H}_{1}\otimes\mathcal{H}_{2}## is defined such that $$\langle\phi ,\psi\vert\psi ,\phi\rangle =\left(\langle\phi\rvert\otimes\langle\psi\lvert\right)\left(\lvert\psi\rangle\otimes\lvert\phi\rangle\right) =\langle\psi\lvert\psi\rangle_{1}\langle\phi\lvert\phi\rangle_{2}$$ where ##\langle\cdot\lvert\cdot\rangle_{1}## is the inner product defined on ##\mathcal{H}_{1}## and ##\langle\cdot\lvert\cdot\rangle_{2}## the inner product defined on ##\mathcal{H}_{2}##.

How though is the outer product defined? Is it simply $$\lvert\psi ,\phi\rangle\langle\phi ,\psi\rvert =\lvert\psi\rangle\otimes\lvert\phi\rangle\langle\phi\rvert\otimes\langle\psi\rvert =\lvert\psi\rangle\langle\psi\rvert_{1} \lvert\phi\rangle\langle\phi\rvert_{2}$$ where ##\lvert\psi\rangle\langle\psi\rvert_{1}## is the outer product in ##\mathcal{H}_{1}## and ##\lvert\phi\rangle\langle\phi\rvert_{2}## is the outer product in ##\mathcal{H}_{2}##.

It might be easier to think of more general outer products, rather than the outer product of an element with itself, as you have written.
That is, consider the outer product ##|a,b\rangle\langle c,d|=|a\rangle\otimes |b\rangle\langle c|\otimes\langle d|##.

This is an operator on the product space ##\mathcal{H}_{1}\otimes\mathcal{H}_{2}##. To see what it does, we apply it to an element of that space ##|e,f\rangle=|e\rangle\otimes |f\rangle##.

That symbol string on the RHS is not intrinsically meaningful. It juxtaposes the two operators ##\lvert a\rangle\langle c\rvert## and ##\lvert b\rangle\langle d\rvert## but:
- since they are not scalars, the juxtaposition cannot be interpreted as scalar multiplication
- since they are operators on different spaces (##\mathcal H_1## and ##\mathcal H_2##), the juxtaposition cannot be interpreted as composition of operation.

Should it be something like $$\lvert a,b\rangle\langle c,d\rvert =\lvert a\rangle\langle c\rvert\otimes\lvert b\rangle\langle d\rvert$$

That is one of a number of intrinsically meaningful ways of writing it. It has the same meaning as what I wrote above ##(
|a\rangle\otimes |b\rangle)(\langle c|\otimes\langle d|)## (I have added parentheses here that were only implied above, to make it clear what operations are being performed).